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The Pauli equation can be obtained by taking the non relativistic limit of the interacting Dirac equation (https://en.wikipedia.org/wiki/Pauli_equation, 2). In the process, you remove a phase of $$e^{-imc^{2}/\hbar}.$$ Starting from the solution to the Dirac equation (positive energy and spin Dirac equation solution) $$\Psi_{\uparrow,\mathbf{p}}^{+}(\mathbf{r},t)=\left(\begin{array}{c} 1 \\ 0 \\ \frac{cp_{z}}{\mathcal{E}_{p}+mc^{2}} \\ \frac{c(p_{x}+ip_{y})}{\mathcal{E}_{p}+mc^{2}} \end{array} \right)e^{i(\mathbf{p}\cdot\mathbf{r}-\mathcal{E}_{p}t)/\hbar}. $$ In the non relativistic limit, $$ \mathcal{E}=mc^{2}\sqrt{1+\frac{p^{2}}{m^{2}c^{2}}}\approx mc^{2}+\frac{p^{2}}{2m}.$$ Hence the overall phase in the Dirac equation can be expressed as follows: $$e^{-i(\textbf{p}\cdot\textbf{r}-\mathcal{E}t)/\hbar}\approx e^{-imc^2/\hbar} \exp\left(\!-\frac i\hbar \left[\textbf{p}\cdot\textbf{r}+\frac{p^2}{2m}\,t\right]\right).$$ In the derivation of the Pauli equation, you cancel the phase explicitly containing the rest mass energy. Since the only explicit time dependence in the wave function after this cancellation is in the phase, which is now expressed in terms of kinetic energy, applying the energy operator of the Pauli equation gives $$i\hbar\,\partial_{t}\psi=\frac{p^{2}}{2m}\,\psi.$$ For a free particle, the energy operator is giving the total energy. However, clearly if there is potential energy, the energy operator still will only return kinetic energy, and not the potential. So, my question is related to solutions to the interacting Dirac equation; what does $\mathcal{E}$ in the phase represent? Should it be contain extra terms, to account for the potential energy?

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In a relativistic context, all energy is kinetic. Potential energies alter the dynamics, which then manifests itself as a change in kinetic energy. The reason is simple: energy is a label that comes from one of the Casimir operators of the Lorentz group ($P_{\mu}P^{\mu}$), and the Lorentz group knows nothing about potential energies. It knows only about movement in space-time (i.e, kinetic energy)

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  • $\begingroup$ Would you consider rest mass energy to be kinetic energy? $\endgroup$
    – Andrew
    Jun 6, 2022 at 3:13
  • $\begingroup$ Yes. Think of it as the kinetic energy associated with time-like geodesics. $\endgroup$ Jun 6, 2022 at 5:14
  • $\begingroup$ So how would one obtain potential energy from the non relativistic limit of the relativistic dispersion relation? ($\mathcal{E}^{2}=m^{2}c^{4}+P^{2}c^{2}$) $\endgroup$ Jun 6, 2022 at 5:49
  • $\begingroup$ It is not possible to do this. In a general quantum field theory, the potential energies felt by the particles can be computed by inserting the propagators of the gauge bosons in the relevant Feynman diagrams. That is, it depends on the dynamics of the theory and it is not captured by the physics of any individual particle. Your confusion seems to come from the fact that energy in classical physics plays a very different role than in a QFT. For example, why can't we solve scattering problems by conservation of energy in QED if we can do it for billiard balls collinding? Think about this. $\endgroup$ Jun 6, 2022 at 7:39
  • $\begingroup$ TL;DR, that is not where the potential energy comes from. $\endgroup$ Jun 6, 2022 at 7:39
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Your formulas do not seem to make sense other than for free Dirac particles. However, you are asking what will happen if, for example, a Dirac particle has potential energy. Then you should start with a solution of the Dirac equation in an external field, for example, electric field.

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